# Ngô Quốc Anh

## July 31, 2009

### A property of the essentially bounded function

Let $E$ be a subset of $\mathbb R^n$ with $|E| < \infty$ in Lebesgue sense. Suppose $f \in L^\infty(E)$ and $\| f\|_\infty > 0$. Set

$\displaystyle {a_n} = \int_E {{{\left| f \right|}^n}}$

for $n=1,2,3,...$ Show that

$\displaystyle\mathop {\lim }\limits_{n \to \infty } \frac{{{a_{n + 1}}}} {{{a_n}}} = {\left\| f \right\|_\infty }$.

Solution. For any $\alpha$ with $0<\alpha < \|f\|_\infty$, let

$\displaystyle{E_\alpha } = \left\{ {x \in E: f\left( x \right) \geqslant \alpha } \right\}$

and

$\displaystyle {F_\alpha } = E\backslash {E_\alpha }$

then $|E_\alpha|>0$. For any $k \in \mathbb N$, by the Dominated Convergence Theorem,

$\displaystyle\mathop{\lim }\limits_{n\to\infty }\left({\dfrac{{\int_{{F_\alpha }}{{{\left| f\right|}^{n+k}}}}}{{\int_{{E_\alpha }}{{{\left| f\right|}^{n}}}}}}\right)\leqslant\underbrace{\mathop{\lim }\limits_{n\to\infty }\frac{1}{{\left|{{E_\alpha }}\right|}}\int_{{F_\alpha }}{{{\left|{\frac{f}{\alpha }}\right|}^{n}}\left\| f\right\|_\infty^{k}}}_{0}$.

Hence

$\displaystyle\mathop{\lim\inf }\limits_{n\to\infty }\left({\frac{{\int_{E}{{{\left| f\right|}^{n+1}}}}}{{\int_{E}{{{\left| f\right|}^{n}}}}}}\right)\geqslant\mathop{\lim\inf }\limits_{n\to\infty }\left({\frac{{\alpha\int_{{E_\alpha }}{{{\left| f\right|}^{n}}}+\int_{{F_\alpha }}{{{\left| f\right|}^{n+1}}}}}{{\int_{{E_\alpha }}{{{\left| f\right|}^{n}}}+\int_{{F_\alpha }}{{{\left| f\right|}^{n}}}}}}\right) =\alpha$.

Letting $\alpha \nearrow {\left\| f \right\|_\infty }$, we get that

$\displaystyle\mathop{\lim }\limits_{n\to\infty }\left({\frac{{\int_{E}{{{\left| f\right|}^{n+1}}}}}{{\int_{E}{{{\left| f\right|}^{n}}}}}}\right) ={\left\| f\right\|_\infty }$.

As an application, if we put $a_0 = 1$, then from

$\displaystyle {a_{n + 1}} = \frac{{{a_1}}} {{{a_0}}}.\frac{{{a_2}}} {{{a_1}}} \cdots\frac{{{a_{n + 1}}}} {{{a_n}}}$

we deduce that

$\displaystyle\mathop{\lim }\limits_{n\to\infty }\sqrt[n]{{{a_{n}}}}=\mathop{\lim }\limits_{n\to\infty }\frac{{{a_{n+1}}}}{{{a_{n}}}}={\left\| f\right\|_\infty }$.

In other words,

$\displaystyle\mathop{\lim }\limits_{n\to\infty }{\left({\int_{E}{{{\left| f\right|}^{n}}}}\right)^{\frac{1}{n}}}={\left\| f\right\|_\infty }$.

### Picard’s Theorem + Hadamard’s Theorem = ?

Question. Let $f$ be an entire non-constant function that satisfies the functional equation

$\displaystyle f(1 - z) = 1 - f(z)$

for all $z \in \mathbb C$. Show that $f(\mathbb C) = \mathbb C$.

Solution. Assume by contradiction, then W.L.O.G. by the Picard’s Theorem we can assume that $f$ misses $a \in \mathbb C$. By the Hadamard’s Theorem,

$\displaystyle f(z)-a = e^{p(z)}$

for some polynomial $p$. Therefore,

$\displaystyle f(z) = a +e^{p(z)}$

for all $z \in \mathbb C$. From the fact that

$\displaystyle f(1-z)=1-f(z)$

we get

$\displaystyle \underbrace{a+{e^{p\left({1-z}\right)}}}_{f\left({1-z}\right)}=\underbrace{1-\left({a+{e^{p\left( z\right)}}}\right)}_{1-f\left( z\right)}$

which yields

$\displaystyle {e^{p\left( z \right)}} = 2a - 1 + {e^{p\left( {1 - z} \right)}}$.

Put $z=0$ and $z=1$, we obtain

$\displaystyle {e^{p\left( 0 \right)}} = 2a - 1 + {e^{p\left( 1 \right)}}, \quad {e^{p\left( 1 \right)}} = 2a - 1 + {e^{p\left( 0 \right)}}$.

Hence

$\displaystyle {e^{p\left( 0 \right)}} = 2\left( {2a - 1} \right) + {e^{p\left( 0 \right)}}$

which implies $a=\frac{1}{2}$. From the identity $f(1-z)=1-f(z)$ put $z=\frac{1}{2}$ we then deduce that

$\displaystyle f\left( {\frac{1} {2}} \right) = \underbrace {\frac{1} {2}}_a$

Corrigendum to the previous proof. Assume by contradiction, then $f$ takes all values except possibly some $a$. Also notethat $f(1/2) = 1/2$. So we can assume $a \neq 1/2$ which means that $1-a \ne a$. It then does take the value $1-a$, and hence takes the value $a$. The proof is now complete.